University of South Carolina Aiken Acidic and Basic Salts Worksheet Answer the following assignment using the information included in the file. Please incl

University of South Carolina Aiken Acidic and Basic Salts Worksheet Answer the following assignment using the information included in the file. Please include your work. Acidic and Basic Conjugate Salts – Their pH in Aqueous Solution
In this lab, as a thought experiment, imagine dissolving sodium chloride (table salt) in water. As
an ionic compound (a metal ion with a nonmetal ion) it dissolves as each of its individual ions—
one sodium ion and one chloride ion according to the formula NaCl. Measuring the pH of this
table salt solution with a calibrated pH meter shows a pH value very close to 7 or neutral pH.
This indicates that those two ions do not disturb the acid/base chemistry in water—in other
words they are acid/base spectator ions. In fact, salt ions that are the cations of any strong base
and the anions from any strong acid (except sulfuric acid) are acid/base spectator ions. That is
why I asked you to remember the list of six strong acids in water and the six common strong
bases in water. Doing so saves time in so many ways. This means that solid salts like KNO3,
BaBr2 or LiClO4 for instance dissolved in water are A/B spectator ions. Look at your list of strong
acids and bases and you will see the cations above associated with the strong bases on that list
and the anions above associated with the strong acids on that list.
If I told you then that a salt like sodium nitrite, NaNO2, when dissolved in water gave a pH of 8.5
to 9.0, you would attribute that change in pH not to the sodium ion but to the nitrite ion, NO2–.
In fact, there are many salts that contain a spectator anion or spectator cation and some other
kind of ion. Often those other ions turn out to change the pH and can be shown to be the
conjugate acid of a weak base or the conjugate base of a weak acid. Those conjugate ions, along
with a spectator ion, when they dissolve in water behave as any weak reacting material does.
They set up an equilibrium that results in a slight excess of either H + ions in the case of a
conjugate acid or a slight excess in OH– ions in the case of a conjugate base. And like any
equilibrium, once we have the balanced reaction, we can set up an ICE scheme for it and
possibly predict the equilibrium concentrations of the materials in the reaction. Because the pH
is disturbed and we have a way to measure that disturbance, a calibrated pH probe, this
experiment becomes a Situation 1 calculation where a measured concentration at equilibrium
and the ICE scheme allows us to experimentally find the equilibrium constant. We will be
obtaining the equilibrium constants for these conjugate acids and bases by measuring the pH of
the various salt solutions. Once we have those conjugate equilibrium constants, we will
calculate the Ka or Kb value for the weak acid or base that gave rise to the conjugate in the first
place and compare those to the textbook values. Remember to save space, and therefore
money, the publishers often just include the Ka or Kb values and leave it up to the reader to find
Kca or Kcb.
Turn to page 153 in your lab manual and look at the first table. We will be focusing most of our
attention here. First you will need to calculate the molar masses for all four salts. Just write
them under the formula given in each box in column 1. Column 2 is the weighed mass of each
salt used in our imagined experiment. Because the procedure calls for 0.40 grams of each, we
will imagine that is what we used. So those are the entries for that column. We are supposed to
dissolve the salt in 50.0 mL of water. We can then calculate an approximate molarity by dividing
the mass of each salt by its molar mass, this gives the number of moles of salt, and dividing that
result by 0.0500 L following the definition of molarity, moles per liter. Column 3 is made up of
the entries of molarity for each salt solution. Column 4 is the last column where experimental
data is provided. The average pH values for each solution based on past experiments are as
follows: NaF ? pH = 8.1; NaC2H3O2 ? pH = 9.1; NH4Cl ? 4.8; NaHCO3 ? 10.1. Place these
values in the appropriate box in column 4. Based strictly on the measured pH you should be
able to decide if the salt behaves as an acidic material or as a base material. Those are the
entries you need to make in column 5.
Now we get down to the nitty gritty—finding the Kca or Kcb for the selection of salts. Table 2 on
page 153 is an example ICE scheme for the salt NaF. We will use it as our example. Obtain the
balanced reaction: F–(aq) + H2O(l) ? HF(aq) + OH–(aq). Table 2 does not include water as it
is the solvent and in dilute solutions we usually exclude the solvent for equilibrium expressions.
We need the full algebraic solution to the ICE scheme. You should be able to see that [F–]o is the
value from Table 1, column 3 and the other materials are zero concentration initially. Using the
change as minus on the LHS and plus on the RHS gives us for the equilibrium concentrations:
[F–]eq = [F–]o – x
[HF]eq = x
[OH–]eq = x
Now we use our old friend the pH Cycle, 14 – pH = pOH and [OH–]eq = x = 10–pOH. With that in
hand, I plug the needed information into the expression and solve for Kcb; Kcb = x2/([F–]o – x).
You do not need to use Table 2 specifically but to do the calculations, you will need to set things
up properly.
Requirements:
1) Fill in Table 1, Columns 1 – 5 as described in an earlier paragraph.
2) Fill in Column 6 by finding the Kcb value for either sodium acetate or sodium bicarbonate but
not both. Because sodium fluoride was our example above, do not include it in your choices.
3) Then everyone should find the Kca value for ammonium chloride. You should be able to find
the balanced reaction in the textbook. The only other point is to realize that NH4+ is a
conjugate acid so it produces H+ ion. We therefore do not need to find the pOH or the [OH–]eq
for the solution. The pH will give us the [H+]eq concentration as 10–pH.
4) Once you have calculated your two K values for the conjugates, determine the K values for
their associated weak acid or weak base using the equations Kw = Ka x Kcb or Kw = Kb x Kca.
Compare the values you got for Ka and Kb for your two salts side-by-side with the values in the
textbook.
Show your work in the lab notebook. You do not need to do any questions on page 154.

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